All equations are always in one frame or, even less confusing, you work with invariant objects, namely the vectors themselves, and not with their components. In this case, let's take ##\vec{e}_j## as the time-independent basis vectors in an inertial reference frame and ##\vec{e}_k'## the time-dependent vectors in the rotating frame. Then there's a rotation matrix ##D_{kj}## such that
$$\vec{e}_k'=D_{kj} \vec{e}_j.$$
Now take any vector ##\vec{A}##. You have
$$\frac{\mathrm{d}}{\mathrm{d} t} \vec{A}=\dot{\vec{A}}=\frac{\mathrm{d}}{\mathrm{d} t} (A_k' \vec{e}_k').$$
Now you have
$$\dot{\vec{A}}=\dot{A}_k' \vec{e}_k' + A_k' \dot{\vec{e}}_k'. \qquad (1)$$
But now
$$\dot{\vec{e}}_k'=\dot{D}_{kj} \vec{e}_j=\dot{D}_{kj} (D^{-1})_{jl} \vec{e}_l'=\epsilon_{klm} \Omega_m' \vec{e}_l', \qquad (2)$$
where I have used that because of
$$\hat{D} \hat{D}^T=\hat{1}$$
the matrix ##\dot{D}_{kj} (D^{-1})_{jl}## is antisymmetric (prove it!). Plugging this in (2) in (1) finally gives
$$\dot{\vec{A}}=\dot{A}_k' \vec{e}_k' + \epsilon_{klm} A_k' \Omega_m' \vec{e}_l = (\dot{A}_l'+\epsilon_{lmk} \Omega_m' A_k') \vec{e}_l'.$$
This means that for the components of ##\dot{\vec{A}}## you get
$$\mathrm{D}_t \underline{A}'=\mathrm{d}_t \underline{A}' + \underline{\Omega}' \times \underline{A}',$$
where the underlined symbols mean the component-column vectors wrt. the basis ##\vec{e}_k'##.